If `CH_(2)COOH+OH^(-)rarrCH_(3)COO^(-)+H_(2)O+q_(1)`
`H^(+)+OH^(-)rarrH_(2)O+q_(2)`
then the enthalpy change for the equal to

Options

A. `(q_(1)+q_(2))`
B. `(q_(1)-q_(2))`
C. `(q_(2)-q_(1))`
D. `-(q_(1)+q_(2))`

Explanation

To find the enthalpy change for the reactions given, we will analyze both reactions step by step.

### Step 1: Understand the reactions
We have two reactions:
1. \( CH_2COOH + OH^- \rightarrow CH_3COO^- + H_2O + q_1 \)
2. \( H^+ + OH^- \rightarrow H_2O + q_2 \)

### Step 2: Identify the type of reactions
- The first reaction involves a weak acid (\( CH_2COOH \)) reacting with a strong base (\( OH^- \)).
- The second reaction is a neutralization reaction between a strong acid (\( H^+ \)) and a strong base (\( OH^- \)).

### Step 3: Define the heat of the reactions
- \( q_1 \) is the heat evolved in the first reaction (weak acid with strong base).
- \( q_2 \) is the heat evolved in the second reaction (strong acid with strong base).

### Step 4: Relate the heats of the reactions
The heat of reaction for the first reaction (\( q_1 \)) is less than that of the second reaction (\( q_2 \)), because the first reaction involves the dissociation of a weak acid.

### Step 5: Write the relationship
The difference in heat between the two reactions can be expressed as:
\[ q_2 - q_1 = \text{Heat of dissociation of the weak acid} \]

### Step 6: Conclusion
Thus, the enthalpy change for the reaction involving the weak acid and the strong base can be expressed as:
\[ \Delta H = q_2 - q_1 \]

### Final Answer
The enthalpy change for the reaction is given by:
\[ \Delta H = q_2 - q_1 \]

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